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There is a procedure called polynomial long division, if you are comfortable with long division of numbers then it isn't too different. You can see an example of the procedure here. However, if you don't recall long division we can rewrite the numerator so that it includes a multiple of the denominator. We do this term by term starting with the leading term (that is, $\\simplify{{r * m}x ^ 3}$).

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\\[\\begin{eqnarray*} \\simplify[std]{{r * m} * x ^ 3 + {n * r + m * s} * x ^ 2 + {n * s + t * r} * x + {t * n + be}}&=&\\simplify[std]{(x+{s})x^2+{n}x^2+{n*s+t}x+{t*n+be}}\\\\&=&\\simplify[std]{(x+{s})x^2+(x+{s})*{n}x+{t}x+{t*n+be}}\\\\ &=&\\simplify[std]{(x+{s})x^2+(x+{s})*{n}x+(x+{s})*{t}+{t*n+be-s*t}}\\\\ &=&\\simplify[std]{(x+{s})(x^2+{n}x+{t})+{t*n+be-s*t}} \\end{eqnarray*} \\]

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Hence
\\[\\frac{\\simplify[std]{{r * m} * x ^ 3 + {n * r + m * s} * x ^ 2 + {n * s + t * r} * x + {t * n + be}}}{\\simplify[std]{{r}x+{s}}}=\\simplify[std]{x^2+{n}x+{t}+{t*n+be-s*t}/({r}x+{s})}\\]

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Dividing a cubic polynomial by a linear polynomial. Find quotient and remainder.

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Input numbers as integers not decimals.

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$q(x)=\\;\\;$[[0]]

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Input all numbers as integers and not as decimals.

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$r=\\;\\;$[[1]]

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Divide $\\displaystyle{\\simplify[std]{{r * m} * x ^ 3 + {n * r + m * s} * x ^ 2 + {n * s + t * r} * x + {t * n + be}}}$ by $\\simplify[std]{{r}x+{s}}$ so that:
\\[\\frac{\\simplify[std]{{r * m} * x ^ 3 + {n * r + m * s} * x ^ 2 + {n * s + t * r} * x + {t * n + be}}}{\\simplify[std]{{r}x+{s}}}=q(x)+\\frac{r}{\\simplify[std]{{r}x+{s}}}\\]

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where $q(x)$ is the quotient polynomial and $r$ is the remainder ($r$ is a constant).

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The coefficients of $q(x)$ are integers, do not input as decimals.

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